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  • 【leetcode】448. Find All Numbers Disappeared in an Array

    题目如下:

    解题思路:本题对时间复杂度和空间复杂度都有要求,特别是空间,所以不能用字典之类的来记录已经出现的值。这里可以采用值-下标映射的方法,即把所有元素移动到其值减1的对应的下标的位置上,移动完成后,下标和值不匹配的元素即为缺失的number。例如输入[4,3,2,7,8,2,3,1],

    [4, 3, 2, 7, 8, 2, 3, 1] # 初始状态
    [7, 3, 2, 4, 8, 2, 3, 1] #4和下标为4-1的元素互换
    [3, 3, 2, 4, 8, 2, 7, 1] #7和下标为7-1的元素互换
    [2, 3, 3, 4, 8, 2, 7, 1] #依次类推
    [3, 2, 3, 4, 8, 2, 7, 1]
    [3, 2, 3, 4, 8, 2, 7, 1]
    [3, 2, 3, 4, 8, 2, 7, 1]
    [3, 2, 3, 4, 8, 2, 7, 1]
    [3, 2, 3, 4, 8, 2, 7, 1]
    [3, 2, 3, 4, 1, 2, 7, 8]
    [1, 2, 3, 4, 3, 2, 7, 8]
    [1, 2, 3, 4, 3, 2, 7, 8]
    [1, 2, 3, 4, 3, 2, 7, 8]
    [1, 2, 3, 4, 3, 2, 7, 8]
    [1, 2, 3, 4, 3, 2, 7, 8] #标红的元素和下标不匹配

    代码如下:

    class Solution(object):
        def findDisappearedNumbers(self, nums):
            """
            :type nums: List[int]
            :rtype: List[int]
            """
            inx = 0
            while inx < len(nums):
                # [4,3,2,7,8,2,3,1]
                if nums[inx] != nums[nums[inx]-1]:
                    #swap 4 and 7, to make val match inx
                    src = nums[inx]  # 4
                    desc = nums[nums[inx]-1] # 7
                    nums[inx] = desc
                    nums[src - 1] = src
                else:
                    inx += 1
            res = []
            for i,v in enumerate(nums):
                if i + 1 != v:
                    res.append(i+1)
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/9303717.html
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